SOLUTION: A hotel has 200 rooms. Those with kitchen facilities rent for $100 per night and those without kitchen facilities rent for $80 per night. On a night when the hotel was completely

Algebra ->  Linear-equations -> SOLUTION: A hotel has 200 rooms. Those with kitchen facilities rent for $100 per night and those without kitchen facilities rent for $80 per night. On a night when the hotel was completely       Log On


   

Question 835514: A hotel has 200 rooms. Those with kitchen facilities rent for $100 per night and those without kitchen
facilities rent for $80 per night. On a night when the hotel was completely occupied, revenues were
$17,000. How many of each type of room does the hotel have?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x count rooms without kitchen
y count rooms WITH kitchen

x%2By=200, because all rooms occupied;
80x%2B100y=17000, that night's revenue.

Simplifying the revenue equation, 8x%2B10y=1700
4x%2B5y=850

Use the system:
----------------------
x+y=200
----------------------
4x+5y=850
----------------------

Try Elimination. Multitply the room count equation by 4, and....